2

Use the exercises below to complete the proof.

Exercises:

1. Check that in Theorems 24 and 25 it su¬ces to prove the case k = 1

and d = 0; the general case then follows through suitable changes of

variables.

2. For a sector S = S(0, ±) of in¬nite radius, let f ∈ A(1) (S, E ). For

su¬ciently large x0 > 0 and t > 0, show

∞

e(t’u)(x0 +iy) ’ e(t’u)(x0 ’iy)

lim f (u) du = f (t), (5.5)

2πi (t ’ u)

y’∞ 0

integrating along the positive real axis; observe that u = t is a re-

movable singularity of the integrand.

5.4 A Di¬erent Representation for Borel Operators

Here we give a di¬erent integral representation for Borel operators of order

k > 1/2, using Mittag-Le¬„er™s function as a kernel. To do so, we ¬rst

prove the following well-known result upon the behavior of Mittag-Le¬„er™s

function as the variable z ’ ∞; for a more complete asymptotic analysis

of Es (z), compare the exercises below.

84 5. Integral Operators

Lemma 6 (Asymptotic of Mittag-Leffler™s function)

Let 0 < s < 2. Then for S’ = S(π, (2 ’ s)π) we have

z Es (z) ’’ ’1/“(1 ’ s), z ’ ∞,

S’ (5.6)

while for S+ = S(0, sπ) we have

Es (z) = s’1 exp[z 1/s ] + Es (z),

˜ (5.7)

˜

with z Es (z) bounded in S+ .

Proof: In the integral representation (B.19) (p. 233) one can deform the

path of integration γ (owing to Cauchy™s integral theorem) such that the

two radial parts follow the two rays arg w = ±(π+µ)/2, for arbitrarily small

µ > 0. Doing so, we ¬nd that for z with |π ’ arg z| ¤ (1 ’ s/2)π ’ sµ the

denominator ws ’ z never vanishes, so that we may interchange integration

and limit, by which we obtain (5.6), owing to (B.10) (p. 228). On the other

hand, for z ∈ S+ we can use the Residue Theorem to show

Es (z) = s’1 exp[z 1/s ] + Es (z),

˜ |z| > 1,

˜

where Es (z) has exactly the same integral representation (B.19), but for

˜

|z| > 1. From this representation, it is easy to see that z Es (z) remains

bounded as z ’ ∞ in S+ . 2

Remark 6: Since exp[z 1/s ] is bounded in the sectors sπ/2 ¤ ± arg z ¤

3sπ/2, we observe that (5.7) remains valid in the sector S(0, 3sπ). This

3

will be used in the proof of the next theorem.

For s as above, consider any sectorial region G = G(d, ±) of opening

± > sπ, and let f ∈ H(G, E ) be continuous at the origin. With k = 1/s

and γk („ ) as in Section 5.2, let

’1 dz

g(u) = Es (u/z) f (z) . (5.8)

2πi z

γk („ )

According to Lemma 6, the integral is absolutely convergent for u with

|d ’ arg u| < (± ’ sπ)/2 and „ ≈ arg u, and convergence is locally uniform

there, so that g(u) is analytic. We now show

Theorem 26 For 0 < s < 2 and any sectorial region G = G(d, ±) of

opening ± > sπ, let f ∈ H(G, E ) be continuous at the origin and de¬ne g

by (5.8). Then we have

g = Bk f, k = 1/s.

5.5 General Integral Operators 85

Proof: The integral for the Borel operator can be written as

’1 dz

k

kf (z) e(u/z)

(Bk f )(u) = .

2πi z

γk („ )

Hence, according to (5.7), resp. Remark 6, we have

’1 dz

˜

g(u) ’ (Bk f )(u) = Es (u/z) .

2πi z

γk („ )

Since the integrand is bounded in the interior of the path of integration,

we can use Cauchy™s integral theorem to conclude g ’ Bk f = 0. 2

The above representation of Borel operators, aside from being interesting

in its own right, also will serve as a starting point for more general integral

operators in the next section. Concerning the restriction of s to the interval

0 < s < 2, so that k = 1/s > 1/2, one should note that this is forced by the

fact that Mittag-Le¬„er™s function behaves di¬erently for other values of s.

In later chapters we shall see other results indicating that the value of 1/2

is special in the sense that some aspects of the theory of summability of

power series change when the parameter k is larger, resp. smaller, than 1/2.

Exercises: Let 0 < s < 2 and S’ = S(π, (2 ’ s)π).

1. For N ≥ 0, show

N ’1

1’N

z ’n /“(1 ’ sn),

Es (z; N ) ’

Es (z) = z (5.9)

n=1

ew wsN ’1 (ws ’ z)’1 dw.

with 2πi Es (z; N ) = γ

2. For Es (z; N ) as above, show z Es (z; N ) ’ ’1/“(1’sN ), as z ’ ∞ in

S’ . Interpret this, together with (5.9), as an asymptotic for Es (1/z).

5.5 General Integral Operators

In this section, we are going to study some generalizations of Laplace resp.

Borel operators. In detail, we are going to de¬ne pairs of integral operators,

one being the inverse of the other on suitable function spaces, and each pair

being associated with one particular “sequence of moments.” To do so, we

introduce pairs of functions serving as kernels for these operators.

Kernel Functions

A pair of C -valued functions e(z), E(z) will be called kernel

functions if for some k > 1/2 the following holds:

86 5. Integral Operators

• The function e(z) is holomorphic in S+ = S(0, π/k), and

z ’1 e(z) is integrable at the origin, meaning that the in-

x

tegral 0 0 x’1 |e(xei„ )| dx exists, for arbitrary x0 > 0 and

2k|„ | < π. Moreover, for every µ > 0 there exist constants

c, K > 0 such that

|e(z)| ¤ c exp[’(|z|/K)k ], 2k| arg z| ¤ π ’ µ. (5.10)

• For positive real z = x, the values e(x) are positive real.

• The function E(z) is entire and and of exponential growth

at most k. Moreover, in S’ = S(π, π(2 ’ 1/k)), i.e., the

complement of S+ in C , the function z ’1 E(1/z) is inte-

grable at the origin in the above sense.

• The functions e(z), E(z) are linked by the following mo-

ment condition: De¬ne the moment function corresponding

to the kernel e(z) by

∞

Re u ≥ 0.

xu’1 e(x) dx,

m(u) = (5.11)

0

Note that the integral converges absolutely and locally

uniformly for these u, so that m(u) is holomorphic for

Re u > 0 and continuous up to the imaginary axis, and

the values m(x) are positive real numbers for x ≥ 0. Then

the power series expansion of E(z) equals

∞

zn

E(z) = . (5.12)

m(n)

0

The number k will sometimes be referred to as the order of the

pair of kernel functions e(z), E(z).

Observe that the kernel E(z) is uniquely determined by e(z) because of

(5.12). Obviously, (5.10) implies k m(n) ¤ c K n “(n/k), n ≥ 1. On the

other hand, by assumption we have that E(z) is of exponential growth at

˜

most k, meaning by de¬nition |E(z)| ¤ c exp[K|z|]k , for su¬ciently large

˜

˜˜

c, K > 0. From the proof of Theorem 69 (p. 233) we conclude that this

ˆˆ ˆˆ

implies existence of c, K > 0 so that m(n) ≥ c K n “(1 + n/k), n ≥ 0. Hence

the moments m(n) are of order “(1+n/k) as n tends to in¬nity in the sense

that [m(n)/“(1 + n/k)]±1/n is bounded. In particular, this shows that the

order of a pair of kernel functions is uniquely de¬ned, and that the (entire)

kernel E(z) is exactly of exponential growth k, or in other words, is of

order k and ¬nite type. As a ¬rst example of such kernel functions we take

e(z) = k z k exp[’z k ]; in this case m(u) = “(1 + u/k) and E(z) = E1/k (z).

Other examples of interest will follow in the exercises and the next section.

5.5 General Integral Operators 87

With the help of any such pair of kernel functions of order k > 1/2, we

now de¬ne a pair of integral operators as follows:

Let S = S(d, ±) be a sector of in¬nite radius, and let f ∈ A(k) (S, E ),

then for |d ’ „ | < ±/2, the integral

∞(„ )

du

(T f )(z) = e(u/z) f (u) (5.13)

u

0

converges absolutely and locally uniformly for z with (4.1) (p. 61), for

su¬ciently large c > 0. As for Laplace operators, a change of „ results in

holomorphic continuation of T f . Hence we conclude that T f is holomorphic

in a sectorial region G(d, ± + π/k).

If G = G(d, ±) is a sectorial region of opening larger than π/k, and

f ∈ H(G, E ) is continuous at the origin, then we de¬ne, with γk („ ) as in

the de¬nition of Borel operators on p. 80,

’1 dz

(T ’ f )(u) = E(u/z) f (z) . (5.14)

2πi z

γk („ )

We conclude as in the proof of Theorem 24 (p. 82) that T ’ f is holomorphic

and of exponential growth at most k in S(d, ± ’ π/k).

In case e(z) = k z k exp[’z k ], the two integral operators coincide with

Laplace resp. Borel operators. Even in general they have many properties

in common with those classical operators, as we now show:

1. For f (u) = u» , with Re » > 0, hence f (u) continuous at the origin,

we have T f (z) = m(») z » ; to see this, make a change of variable in

(5.13) and use (5.11).

∞

2. For f (u) = 0 fn un being entire and of exponential growth at most

k, the function T f is holomorphic for |z| < ρ, with su¬ciently small

∞

ρ > 0, and (T f )(z) = 0 fn m(n) z n , |z| < ρ; to see this, check that

termwise integration of the power series expansion of f is justi¬ed.

3. For w = 0 and z = 0 so that |z/w| is su¬ciently small, it follows from

the above that

∞(„ )

w du

= e(u/z) E(u/w) . (5.15)

w’z u

0

This formula extends to values w = 0 and z = 0 for which both

sides are de¬ned. In particular, this is so for arg w = arg z modulo

2π, since then we can choose „ so that |„ ’ arg z| < π/(2k) and

|π ’ „ + arg w| < π(2 ’ 1/k), implying absolute convergence of the

integral, according to the properties of kernel functions.

4. For a sectorial region G = G(d, ±) of opening more than π/k, and

f ∈ H(G, E ) continuous at the origin, the composition h = T —¦ T ’ f

88 5. Integral Operators

is de¬ned. Interchanging the order of integration and then evaluating

the inner integral, using (5.15), implies

’1 f (w)

h(z) = dw = f (z),

w’z

2πi γk („ )

since γk („ ) has negative orientation. Hence we conclude that T ’ is

an injective integral operator, and T is its inverse. Note, however,

that we do not yet know that either operator is bijective; this will be

shown in Theorem 30 (p. 92).

5. For Re » > 0 and f (z) = z » , we conclude from (5.14) by a change of

variable u/z = w, and using Cauchy™s theorem to deform the path of

integration: